= Solution
Take $n$ to be a positive even integer, as required by <composite Simpson's rule>. There are $n/2$ two-strip panels, each of width $2/n$. Applying the supplied panel bound with the global fourth-<derivative> minimum $0$ and maximum $1$, and summing, gives
$$
\boxed{0\le E_n\le\frac{n}{2}\frac{(2/n)^5}{2880}
=\frac1{180n^4}.}
$$
Here $E_n$ is the nonnegative <absolute error>. In this example <Simpson's rule> overestimates the <integral>, so $E_n=S_n-I$. Each panel has that same sign because its fourth <derivative> is nonnegative; cancellation is not being assumed.<NIST Digital Library of Mathematical Functions/3.5 Quadrature, subsection (ii) Simpson’s Rule>[§3.5(ii), equations 3.5.6–3.5.8: elementary and composite fourth-<derivative> remainders]{ref}
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